*On Jan 16, 2012, at 11:50 PM, Paul Zielinski wrote:*

*PZ: Well if your calculation here is correct, this does appear to put things in a different light.*

JS: Well where is the error? Are you saying this is an original way of looking at an old problem?

*On Jan 16, 2012, at 8:55 PM, Paul Zielinski wrote: Yes I think it's important to emphasize that GCT invariance is sufficient but not necessary for frame invariance. the only reliable invariant for light is ds = 0 no "speed" there This simply means that the time intervals read by clocks moving along null geodesics are all zero.*

Obviously, but that's not THE POINT! Also no material clocks can do that.

If the actual time measured by a clock is

dT(LNIF) = g00^1/2dt(LIF)

Then

ds(invariant) = c(LIF)dT(LNIF)[1 + c(LIF)^-1(g0i/g00)(dx^i/dt) + c(LIF)^-2gij(dx^i/dt)(dx^j/dt)]^1/2

or

GAMMA'ds(invariant) = c(LIF)dT(LNIF)

GAMMA' = [1 + c(LIF)^-1(g0i/g00)(dx^i/dt) + c(LIF)^-2(gij/g00)(dx^i/dt)(dx^j/dt)]^-1/2

---> [1 - (v/c)^2]^1/2 in the EEP tetrad mapping of LNIF to COINCIDENT LIF

for a light ray

ds = 0

1 + c(LIF)^-1(g0i/g00)(dx^i/dt) + c(LIF)^-2(gij/g00)(dx^i/dt)(dx^j/dt) = 0

therefore the PHYSICAL SPEED OF LIGHT =/= COORDINATE SPEED OF LIGHT

1 + g00^-1/2

**A.c**

**'**(LNIF)/c(LIF) - (c(LNIF)/c(LIF))^2 = 0

because (dL(LNIF))^2 = -gijdx^idx^j Lorentzian manifold

(dT)^2 = g00(dt)^2

c(LNIF) = dL(LNIF)/dT(LNIF)

So when

**A**= 0 there is no shift in the observable speed of light in vacuum.